Find the value of ${sin}^{5} θ$

\frac{1}{32} (sin 5 θ - 5 sin 3 θ + 10 sin θ)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{16} (sin 5 θ - 5 sin 3 θ + 10 sin θ)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

^{\frac{1}{16} (sin 5 θ + 5 sin 3 θ + 10 sin θ)}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

^{\frac{1}{32} (sin 5 θ - 5 sin 3 θ + 10 sin θ)}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{1}{16} (sin 5 θ - 5 sin 3 θ + 10 sin θ)$
Let
$x = cos θ + i sin θ$ , $x^{n} = cos n θ + i sin n θ$ ...(By Demovier's thm)
$\Rightarrow \frac{1}{x} = cos θ - i sin θ$
Therefore
$x - \frac{1}{x} = 2 i sin θ$ And $x^{n} - \frac{1}{x^{n}} = 2 i sin n θ$

Therefore
$(x - \frac{1}{x})^{5} = 32 i {sin}^{5} θ$

$x^{5} - 5 x^{3} + 10 x - \frac{10}{x} + \frac{5}{x^{3}} - \frac{1}{x^{5}} = 32 i {sin}^{5} θ$

$(x^{5} - \frac{1}{x^{5}}) - 5 (x^{3} - \frac{1}{x^{3}}) + 10 (x - \frac{1}{x}) = 32 i {sin}^{5} θ$

$2 i [sin 5 θ - 5 sin 3 θ + 10 sin θ] = 32 i {sin}^{5} θ$

$\frac{1}{16} (sin 5 θ - 5 sin 3 θ + 10 sin θ) = {sin}^{5} θ$ .

Suggest Corrections

Similar questions

{(x - 20)}^{2} + {(y - 20)}^{2} = r^{2}

\frac{x^{2}}{4} + \frac{y^{2}}{1} = 100

(20 cos θ, 10 sin θ)

sin 5 θ = 16 {sin}^{5} θ - 20 {sin}^{3} θ + 5 sin θ

(5, 0)

(10 cos θ, 10 sin θ)

2 : 3

P

P

P (x, y)

A (5, 0)

B (10 cos θ, 10 sin θ)

2 : 3,

Prove that following identities:

$s i n 5 θ = 5 s i n θ - 20 s i n^{3} θ + 16 s i n^{5} θ$

Join BYJU'S Learning Program